Another hard combinatorics problem

In how many ways can the elements of [n] be permuted if 1 is to precede both 2 and 3?

My solution: {1, 2, 3} can be permuted in 6 ways. Out of them, only 2: (1, 2, 3) and (1, 3, 2) are acceptable.
By symmetry, all the 6 permutations must occur equal number of times. Total number of permutations is n! So, the acceptable number of permutations is n!/3.